Question: What is the extraneous solution to these equations? $\dfrac{x^2 - 14}{x + 8} = \dfrac{-12x - 46}{x + 8}$
Explanation: Multiply both sides by $x + 8$ $ \dfrac{x^2 - 14}{x + 8} (x + 8) = \dfrac{-12x - 46}{x + 8} (x + 8)$ $ x^2 - 14 = -12x - 46$ Subtract $-12x - 46$ from both sides: $ x^2 - 14 - (-12x - 46) = -12x - 46 - (-12x - 46)$ $ x^2 - 14 + 12x + 46 = 0$ $ x^2 + 32 + 12x = 0$ Factor the expression: $ (x + 4)(x + 8) = 0$ Therefore $x = -4$ or $x = -8$ At $x = -8$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -8$, it is an extraneous solution.